Answer
The function has a maximum value of $11$.
Work Step by Step
RECALL:
(1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens:
(i) upward when $a \gt0$ and has its vertex as its minimum; or
(ii) downward when $a\lt0$ and has its vertex as its maximum.
(2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$
Compare $f(x)=-2x^2+8x+3$ to $f(x)=ax^2+bx+c$.
It can be seen that $a=-2$, $b=8$, $c=3$.
$a\lt0$, hence the graph opens down, hence it's vertex is a maximum.
The maximum value is at $x=-\frac{b}{2a}=-\frac{8}{2\cdot(-2)}=2.$
Thus, the maximum value is $f(2)=-2\cot(2)^2+8\cdot (2)+3=11.$
